我的問題是選擇值,該值是
<option selected="selected" disabled="disabled">Select a city...</option>
的問題是:儘管沒有值這個選項,PHP代碼認爲其值作爲*'選擇一個城市'*。所以,即使沒有選擇一個城市,我的代碼if(isset($_POST['city'])
也會一直返回。我該如何解決這個問題?<選項選擇=「選擇」>文本</option>返回總是儘管不具有值的值
這裏是HTML代碼:
<form id="city_form">
<select id="city" name="city">
<option selected="selected" disabled="disabled">Select a city...</option>
<option value=1 name="city1">City 1</option>
<option value=2 name="city2">City 2</option>
<option value=3 name="city3">City 3</option>
<option value=4 name="city4">City 4</option>
</select>
<input type="submit" value="Submit form" id="submit_button" />
</form>
<div id="show_city"></div>
這裏是的jquery代碼:
$('#submit_button').click(function(e) {
e.preventDefault();
var data = $('#city_form').serialize();
$.post(
"fetch_city.php",
data,
function (response) {
$('#show_city').html(response);
}
);
});
這裏是fetch_city.php:
if (isset($_POST['city']) && !empty($_POST['city'])) {
$city_id = $_POST['city']);
echo $city_id;
} else {
echo "You did not select a city...!";
}
fyi'!empty'意味着'isset',所以如果你測試'!empty' – ThiefMaster
和$ city_id =(int)$ _POST ['city']),你就不會'isset'。 –